1. Given data

• Mass of sphere: $m = 500\text{ g} = 0.5\text{ kg}$
• Speed of centre: $v = 5.00\text{ cm/s} = 0.05\text{ m/s}$
• The sphere is uniform and rolling without slipping.

2. Kinetic energy of a rolling body

For rolling without slipping, $K = K_{\text{trans}} + K_{\text{rot}}$

So, $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

For a uniform solid sphere, moment of inertia about its centre is $I = \frac{2}{5}mR^2$

And for rolling without slipping, $\omega = \frac{v}{R}$

3. Substitute into the rotational part

$K_{\text{rot}} = \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2$

$K_{\text{rot}} = \frac{1}{2}\cdot \frac{2}{5}m v^2 = \frac{1}{5}mv^2$

Thus total kinetic energy is $K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

4. Put the values

First, $v^2 = (0.05)^2 = 0.0025$

Then, $K = \frac{7}{10}(0.5)(0.0025)$

$K = 0.7 \times 0.5 \times 0.0025$

$K = 0.35 \times 0.0025 = 0.000875\text{ J}$

$K = 8.75 \times 10^{-4}\text{ J}$

5. Match with options

$8.75 \times 10^{-4}\text{ J}$ corresponds to Option C.

6. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.